Buck Regulator

Discussion in 'Homework Help' started by fantabulous68, May 20, 2010.

  1. fantabulous68

    Thread Starter Active Member

    Nov 3, 2007
    51
    4
    im stuck on question a iii).
    Somebody guide me please.

    i know Vout=Vin (ton/T)

    f=1/T

    but i dont think the answer is that simple for 7 marks..............
     
  2. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    It's to do with the switching time of the transistor. For your 40V input case what duty cycle do you need? How fast can the transistor reliably switch fully off and on? What would that make the maximum frequency?
     
  3. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    Let's make a trade... I want to know the answers to the first two questions, about why the BJT takes so long to turn off, and how to fix that. I don't know much about BJTs, as I work with FETs much more.

    My understanding of part iii is that the minimum "on" time is 8 uS.

    Vout = Vin * Don where Don is on-state duty cycle, so for the whole input range...

    5 = 40 * Don ... Don = 1/8
    and
    5 = 10 * Don ... Don = 1/2

    The shortest possible time on is 8 microseconds, so the shortest period at 1/8 on is 64 us, so the highest frequency is 15.625 kHz -- however, as an electrical engineer, I would give at least a 100% safety margin, so I would not try to run it above 7.5 kHz... who knows what the actual state of the BJT is in the transition period. Not me... that's why I asked the questions above... please answer them if you can.
     
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  4. fantabulous68

    Thread Starter Active Member

    Nov 3, 2007
    51
    4
    a i) when the power bjt gets base drive then its base region gets filled with charge carriers. For the power BJT to turn off, the charge from the base must be removed. It gets removed by recombination and drift which is a slow process and causes a delay in turning off.:)

    aii) place a resistor across the B-E junction of the power BJT. The resistor provides a discharge path. It draws current away from the base of the power bjt hence drawing charge from the base and discharging the base on turn-off. The current of the resistor can be 10% of the base current of the power bjt. The resistor can be small but not too small because at high currents it will pop.:)
     
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  5. fantabulous68

    Thread Starter Active Member

    Nov 3, 2007
    51
    4
    The min off time is 8 uS not the min on time, right?
     
    Last edited: May 20, 2010
  6. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    What I read is that it takes minimum 8 us to turn off, so the min on time is 8 us.
     
  7. fantabulous68

    Thread Starter Active Member

    Nov 3, 2007
    51
    4
    Yes you are right:) Thanks for helping
     
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